Question

A battery of emf 4 volt and internal resistance $1\Omega$ is connected in parallel with another battery of emf 1 V and internal resistance $1\Omega$ (with their like poles connected together). The combination is used to send current through an external resistance of $2\Omega$. Calculate the current through the external resistance.

Answer 1

Given : $E_1=4V,r_1=1\Omega ,E_2=1V$

$r_2=1\Omega ,R=2\Omega$

To Find:

Current (I) and direction of I

Formula: i. $\sum I=0$ at any junction

ii. $\sum IR=\sum E$

Calculation:

From formula (i) at junction B we get

$I-I_1+I_2$ ............(1)

From the formula for the loop ABCDA we get,

$I\times I_1-I\times I_2=4-1$

$\therefore I_1-I_2=3$ .....(2)

From formula for the loop AEFDA we get,

$I\times I_2-2\times I=4$

$\therefore I\times I_1+2\times (I_1+I_2)=4$

$I_1+2I_1+2I_2=4$

$3I_1+2I_2=4$ ...(3)

Adding [2 $\times$ eq. (2) to eq (3)]

$3I_1+2l_2+2(l_1-l_2)=4+6$

$5l_1=10$

$I_1=2$

Substituting in eq. (2) we get,

$2-I_2=3$

$I_2=1$

$I=I_1+I_2=2+(-1)=1A$